Overview of Elasticity of Materials/Example Problems

Example 1

Problem Statement:

Given a point on an elastic body with stress state $\mathit{\sigma }$, determine the principal stresses and the angle between the original orientation and the orientation with maximum shear stress. Solve this problem both using the analytical equations and using Mohr's circle construction. Stresses are given in MPa.

$\mathit{\sigma }=\left(\begin{array}{ccc}72& 21& 0\\ 21& 32& 0\\ 0& 0& 0\end{array}\right)$

Solution:

We're going to solve this by two methods. First using the analytic solution then using Mohr's circle.

Analytic Solution

We have the equations

$\begin{array}{c}\sigma {\text{'}}_{11}=\frac{{\sigma }_{11}+{\sigma }_{22}}{2}+\frac{{\sigma }_{11}-{\sigma }_{22}}{2}cos\left(2\theta \right)+{\sigma }_{12}sin\left(2\theta \right)\\ \sigma {\text{'}}_{22}=\frac{{\sigma }_{11}+{\sigma }_{22}}{2}-\frac{{\sigma }_{11}-{\sigma }_{22}}{2}cos\left(2\theta \right)-{\sigma }_{12}sin\left(2\theta \right)\end{array}$

and

$\begin{array}{c}\sigma {\text{'}}_{12}=\frac{{\sigma }_{22}-{\sigma }_{11}}{2}sin\left(2\theta \right)+{\sigma }_{12}cos\left(2\theta \right)\end{array}$

where the prime stresses are in the rotated reference frame. When $\sigma {\text{'}}_{12}=0$ the system is in its principal orientation and $\sigma {\text{'}}_{11}={\sigma }_{p1}$ and $\sigma {\text{'}}_{22}={\sigma }_{p2}$. Setting $\sigma {\text{'}}_{12}=0$ and solving yields

$tan\left(2\theta \right)=\frac{2{\sigma }_{12}}{{\sigma }_{22}-{\sigma }_{11}}$

and upon substituting the stresses given in this problem we find $\theta =23.2$°. Substituting this value of $\theta$ and the given values for $\mathit{\sigma }$ yields $\sigma {\text{'}}_{11}={\sigma }_{p1}=81$ MPa $\sigma {\text{'}}_{22}={\sigma }_{p2}=23$ MPa and $\sigma {\text{'}}_{12}=0$ MPa, which verifies that this is indeed the principal axis.

If we are given a stress state that is the principal orientation then in our rotated reference frame above ${\mathit{\sigma }}^{\prime }$ the ${\sigma }_{12}$ term is zero. We find the direction with maximum shear by taking the derivative of $\sigma {\text{'}}_{12}$, setting it to zero, and solving for $\theta$.

$\frac{d\sigma {\text{'}}_{12}}{d\theta }=\frac{d}{d\theta }\left(\frac{{\sigma }_{22}-{\sigma }_{11}}{2}sin\left(2\theta \right)\right)=({\sigma }_{22}-{\sigma }_{11})cos\left(2\theta \right)$

and the maximum occurs when $\theta =45$°. This means the maximum or minimum occurs at $23.2+45=68.2$°. Substituting this into our equations along with $\mathit{\sigma }$ yields $\sigma {\text{'}}_{11}={\sigma }_{p1}=52$ MPa, $\sigma {\text{'}}_{22}={\sigma }_{p2}=52$ MPa and $\sigma {\text{'}}_{12}=-29$ MPa, the minimum. By the symmetry of the system rotating an additional $90$° yields the maximum, $\sigma {\text{'}}_{11}={\sigma }_{p1}=52$ MPa, $\sigma {\text{'}}_{22}={\sigma }_{p2}=52$ MPa, and $\sigma {\text{'}}_{12}=29$ MPa.

Mohr's Circle Solution

Diagram showing the Mohr's circle representation of this example problem.

Following the diagram given here,

$Radius=AB=BC=\sqrt{B{D}^2+C{D}^2}$

and

$tan\left(2\theta \right)=\frac{CD}{BD}$

Given $CD=21$, $OD=72$, and $OE=32$, we concluded that $AD=40$ and $EB=BD=20$. Therefore, $Radius=29$ and $\theta =23.2$°.From the principal orientation$P1=OB+Radius=OE+EB+Radius=81$ MPa and $P2=P1-2Radius=23$ MPa. The minimum and maximum shear stress is $\pm Radius=\pm 29$ MPa and are found by rotating the system $23.2+45=68.2$° and $23.2+45+90=158.2=-21.8^{\circ }$.

Example 2

Problem Statement:

Given a point on an elastic body with stress state $\mathit{\sigma }$, determine the principal stresses. Stresses are given in MPa. [Hint: There is only one way to draw the Mohr's circle representation. Use this to simplify your work. You'll find that there is almost no math once you draw the picture.]

$\mathit{\sigma }=\left(\begin{array}{ccc}0& 6& 8\\ 6& 0& 10\\ 8& 10& 0\end{array}\right)$

Example 3

Problem Statement:

Given a point on an elastic body with stress state $\mathit{\sigma }$, and that rotating about $x_3$ yields stress state ${\mathit{\sigma }}^{\prime }$, fully determine both stress states and the unknown parameter $q$. Stresses are given in MPa.

$\mathit{\sigma }=\left(\begin{array}{ccc}25q& {\sigma }_{12}& 0\\ {\sigma }_{21}& 5q& 0\\ 0& 0& 0\end{array}\right){\mathit{\sigma }}^{\prime }=\left(\begin{array}{ccc}50& 5& 0\\ 5& \sigma {\text{'}}_{22}& 0\\ 0& 0& 0\end{array}\right)$

Solution:

Diagram showing the Mohr's circle representation of this example problem. The shaded regions are similar triangles.

We have the equations

$\begin{array}{c}\sigma {\text{'}}_{11}=\frac{{\sigma }_{11}+{\sigma }_{22}}{2}+\frac{{\sigma }_{11}-{\sigma }_{22}}{2}cos\left(2\theta \right)+{\sigma }_{12}sin\left(2\theta \right)\\ \sigma {\text{'}}_{22}=\frac{{\sigma }_{11}+{\sigma }_{22}}{2}-\frac{{\sigma }_{11}-{\sigma }_{22}}{2}cos\left(2\theta \right)-{\sigma }_{12}sin\left(2\theta \right)\end{array}$

and

$\begin{array}{c}\sigma {\text{'}}_{12}=\frac{{\sigma }_{22}-{\sigma }_{11}}{2}sin\left(2\theta \right)+{\sigma }_{12}cos\left(2\theta \right)\end{array}$

where the prime stresses are in the rotated reference frame.

This immediately allows us to substitute for $\sigma {\text{'}}_{12}$ and $\theta$ to find $q=-1/2$. This give ${\sigma }_{11}=-12.5$ MPa and ${\sigma }_{22}=-2.5$. The invariant relation

${\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33}=I_1$

allows

${\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33}=\sigma {\text{'}}_{11}+\sigma {\text{'}}_{22}+\sigma {\text{'}}_{33}$

with substitutions determines $\sigma {\text{'}}_{22}=-65$ MPa. At this point all the parameters are determined except ${\sigma }_{12}$ and simply substituting into either the $\sigma {\text{'}}_{11}$ or $\sigma {\text{'}}_{22}$ allows us to find ${\sigma }_{12}=57.5$ MPa. The resulting stress tensors, in units of MPa, are

$\mathit{\sigma }=\left(\begin{array}{ccc}-12.5& 57.5& 0\\ 57.5& -2.5& 0\\ 0& 0& 0\end{array}\right){\mathit{\sigma }}^{\prime }=\left(\begin{array}{ccc}50& 5& 0\\ 5& -65& 0\\ 0& 0& 0\end{array}\right)$

The Mohr's circle representation of this solution is shown here. Note that the shaded triangles are similar and therefore can be used to simplify the solution if sought graphically.

Example 4

Problem Statement:

Given a point on an elastic body with stress state $\mathit{\sigma }$, determine the principal stresses and the angle between the original orientation and the orientation with maximum shear stress. Stresses are given in MPa.

$\mathit{\sigma }=\left(\begin{array}{ccc}18& 12& 9\\ 12& 12& -6\\ 9& -6& 6\end{array}\right)$

Example 5

Problem Statement:

Both mathematically and with words explain the impact of applying the transformation tensors ${𝑻}_1$, ${𝑻}_2$,${𝑻}_3$, and ${𝑻}_4$ on vector $𝒂$. Apply the same transformation to rank two tensor $𝒁$.

${𝑻}_1=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right){𝑻}_2=\left(\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right){𝑻}_3=\left(\begin{array}{ccc}-1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right){𝑻}_4=\left(\begin{array}{ccc}cos\left(\theta \right)& 0& sin\left(\theta \right)\\ 0& 1& 0\\ -sin\left(\theta \right)& 0& cos\left(\theta \right)\end{array}\right)$

$𝒂=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)𝒁=\left(\begin{array}{ccc}{Z}_{11}& Z_{12}& Z_{13}\\ Z_{21}& Z_{22}& Z_{23}\\ Z_{31}& Z_{32}& Z_{33}\end{array}\right)$

Solution:

Remembering the definition of tensor multiplication and the implicit summations used in Einstein notation, we know that the rank two $𝑻$ tensor acting on the vector $𝒂$ will necessarily result in terms that look like

$a{\text{'}}_2=a_1{T}_{11}+a_2{T}_{12}+a_3{T}_{13}$

which is similar to "normal" matrix multiplication as you've seen before. In contrast applying the rank two transformation tensor on the rank two $𝒁$ tensor requires a double sum which results in terms that look similar to

$Z{\text{'}}_{12}=T_{11}{T}_{21}{Z}_{11}+T_{11}{T}_{22}{Z}_{12}+T_{11}{T}_{23}{Z}_{13}+T_{12}{T}_{21}{Z}_{21}+T_{12}{T}_{22}{Z}_{22}+T_{12}{T}_{23}{Z}_{23}+T_{13}{T}_{21}{Z}_{31}+T_{13}{T}_{22}{Z}_{32}+T_{13}{T}_{23}{Z}_{33}$

When working with these type of 9-term sums it is usually advisable to use a software package to simply the work. The results are given here.

Transformation tensor 1 results in

${𝒂}^{\prime }=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right){𝒁}^{\prime }=\left(\begin{array}{ccc}{Z}_{11}& Z_{12}& Z_{13}\\ Z_{21}& Z_{22}& Z_{23}\\ Z_{31}& Z_{32}& Z_{33}\end{array}\right)$

which corresponds to the identity transformation, i.e., leave the tensors unmodified.

Transformation tensor 2 results in

${𝒂}^{\prime }=\left(\begin{array}{c}-a_1\\ -a_2\\ -a_3\end{array}\right){𝒁}^{\prime }=\left(\begin{array}{ccc}{Z}_{11}& Z_{12}& Z_{13}\\ Z_{21}& Z_{22}& Z_{23}\\ Z_{31}& Z_{32}& Z_{33}\end{array}\right)$

This transformation is the inversion transformation. This can be seen in the behavior of the vector. Interestingly enough it leaves $𝒁$ unmodified.

Transformation tensor 3 results in

${𝒂}^{\prime }=\left(\begin{array}{c}-a_1\\ a_2\\ -a_3\end{array}\right){𝒁}^{\prime }=\left(\begin{array}{ccc}{Z}_{11}& -Z_{12}& Z_{13}\\ -Z_{21}& Z_{22}& -Z_{23}\\ Z_{31}& -Z_{32}& Z_{33}\end{array}\right)$

This transformation involves mirroring across the $x_1$ and $x_3$ directions. This is equivalent to rotation around the $x_2$-axis by $180$°.

Transformation tensor 4 results in $${𝒂}^{\prime }=\left(\begin{array}{c}{a}_{1}cos\left(\theta \right)+a_{3}sin\left(\theta \right)\\ a_2\\ a3cos\left(\theta \right)-a_{1}sin\left(\theta \right)\end{array}\right){𝒁}^{\prime }=\left(\begin{array}{ccc}{Z}_{11}cos{\left(\theta \right)}^2+Z_{13}cos\left(\theta \right)sin\left(\theta \right)+Z_{31}cos\left(\theta \right)sin\left(\theta \right)+Z_{33}sin{\left(\theta \right)}^2& Z_{12}cos\left(\theta \right)+Z_{32}sin\left(\theta \right)& Z_{13}cos{\left(\theta \right)}^2-Z_{11}cos\left(\theta \right)sin\left(\theta \right)+Z_{33}cos\left(\theta \right)sin\left(\theta \right)-Z_{31}sin{\left(\theta \right)}^2\\ Z_{21}cos\left(\theta \right)+Z_{23}sin\left(\theta \right)& Z_{22}& Z_{23}cos\left(\theta \right)-Z_{21}sin\left(\theta \right)\\ Z_{31}cos{\left(\theta \right)}^2-Z_{11}cos\left(\theta \right)sin\left(\theta \right)+Z_{33}cos\left(\theta \right)sin\left(\theta \right)-Z_{13}sin{\left(\theta \right)}^2& Z_{32}cos\left(\theta \right)-Z_{12}sin\left(\theta \right)& Z_{33}cos{\left(\theta \right)}^2-Z_{13}cos\left(\theta \right)sin\left(\theta \right)-Z_{31}cos\left(\theta \right)sin\left(\theta \right)+Z_{11}sin{\left(\theta \right)}^2\end{array}\right)$$ This transformation involves rotation around the $x_2$-axis. Substituting $180$° for $\theta$° yields identical results to transformation tensor 3.

Example 6

Problem Statement:

Consider the stress state $\mathit{\sigma }$. Write a transformation tensor that rotates the reference frame into the principal directions. Stresses are given in MPa.

$\mathit{\sigma }=\left(\begin{array}{ccc}72& 21& 0\\ 21& 32& 0\\ 0& 0& 0\end{array}\right)$

Solution:

We know from Example 1 that the solution is to rotate by $\theta =23.2$ around the $x_3$ axis and we know the form of the rotation transformation matrix from Example 5, therefore the solution is

$\mathit{\sigma }=\left(\begin{array}{ccc}cos\left(\theta \right)& sin\left(\theta \right)& 0\\ sin\left(\theta \right)& cos\left(\theta \right)& 0\\ 0& 0& 0\end{array}\right)$

where $\theta =23.2$.

Example 7

Problem Statement:

Given the displacement tensor $𝒆$, identify the rotation tensor and strain tensor.

$𝒆=\left(\begin{array}{ccc}0.001& 0.001& 0.002\\ -0.001& 0.002& 0.002\\ 0.001& 0.003& 0.002\end{array}\right)$

Solution:

We can break any tensor into a fully symmetric and fully anti-symmetric tensor resulting in an anti-symmetric rotation tensor

$\mathit{\omega }=\left(\begin{array}{ccc}0& \frac{e_{12}-e_{21}}{2}& \frac{e_{13}-e_{31}}{2}\\ \frac{e_{21}-e_{12}}{2}& 0& \frac{e_{23}-e_{32}}{2}\\ \frac{e_{31}-e_{13}}{2}& \frac{e_{32}-e_{23}}{2}& 0\end{array}\right)=\left(\begin{array}{ccc}0& 0.001& 0.0005\\ -0.001& 0& -0.0005\\ -0.0005& 0.0005& 0\end{array}\right)$

and symmetric strain tensor

$\mathit{\epsilon }=\left(\begin{array}{ccc}{e}_{11}& \frac{e_{12}+e_{21}}{2}& \frac{e_{13}+e_{31}}{2}\\ \frac{e_{21}+e_{12}}{2}& e_{22}& \frac{e_{23}+e_{32}}{2}\\ \frac{e_{31}+e_{13}}{2}& \frac{e_{32}+e_{23}}{2}& e_{33}\end{array}\right)=\left(\begin{array}{ccc}0.001& 0& 0.0015\\ 0& 0& 0.0025\\ 0.0015& 0.0025& 0\end{array}\right)$

resulting ultimately in $𝒆=\mathit{\omega }+\mathit{\epsilon }$.

Example 8

Problem Statement:

Within linear, homogeneous, isotropic elasticity theory, for a given stress state (units in MPa) determine the strain state given the Poisson ratio is 0.40 and the shear modulus is 50 GPa. Identify the hydrostatic stress, deviatoric stresses, strain dilatation, and strain deviator.

$\mathit{\sigma }=\left(\begin{array}{ccc}10& 2& 2\\ 2& 5& -3\\ 2& -3& -5\end{array}\right)$

Solution:

We're given the stress, so returning a solution requires determining the strain, which can be determined by

${\epsilon }_{ij}=\frac{1+\nu }{E}{\sigma }_{ij}-\frac{\nu }{E}{\sigma }_{kk}{\delta }_{ij}$

We're given the Poisson ratio, $\nu$ and shear modulus $G$ but to use this equation we need the elastic modulus $E$. This can be found from

$G=\frac{E}{2(1-\nu )}$

which can be inverted to

$E=2G(1-\nu )$

Upon substitution this yields $E=60$ GPa. The resulting strain tensor is

$\mathit{\epsilon }=\left(\begin{array}{ccc}\frac{1+\nu }{E}{\sigma }_{11}-\frac{\nu }{E}({\sigma }_{22}+{\sigma }_{33})& \frac{{\sigma }_{12}}{2G}& \frac{{\sigma }_{13}}{2G}\\ \frac{{\sigma }_{21}}{2G}& \frac{1+\nu }{E}{\sigma }_{22}-\frac{\nu }{E}({\sigma }_{11}+{\sigma }_{33})& \frac{{\sigma }_{23}}{2G}\\ \frac{{\sigma }_{31}}{2G}& \frac{{\sigma }_{32}}{2G}& \frac{1+\nu }{E}{\sigma }_{33}-\frac{\nu }{E}({\sigma }_{11}+{\sigma }_{22})\end{array}\right)=\left(\begin{array}{ccc}0.1667& 0.0467& 0.0467\\ 0.0467& 0.0500& -0.0700\\ 0.0467& -0.0700& -0.1833\end{array}\right).$

The hydrostatic stress is

${\sigma }_m=\frac{{\sigma }_{kk}}{3}=\frac{{\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33}}{3}=\frac{10+5-5}{3}=3.33\text{ GPa}$

The strain dilatation is

$\mathrm{\Delta }=(1+{\epsilon }_{11})(1+{\epsilon }_{22})(1+{\epsilon }_{33})-1=(1+0.1667)(1+0.0500)(1+0.1833)-1=0.4496$

which results in a mean strain of

${\epsilon }_m=\frac{\mathrm{\Delta }}{3}=\frac{0.4496}{3}=0.1499$

(Note that when strain is small $\mathrm{\Delta }\approx {\epsilon }_{11}+{\epsilon }_{22}+{\epsilon }_{33}=0.400$, which yields ${\epsilon }_m\approx 0.1333$. Unfortunately, in this example the strain is relatively large.) The deviatoric stress and strain are then determined by subtracting the mean stress and strain from their respective tensors diagonal.

${\mathit{\sigma }}^{\prime }=\left(\begin{array}{ccc}10-3.33& 2& 2\\ 2& 5-3.33& -3\\ 2& -3& -5-3.33\end{array}\right)=\left(\begin{array}{ccc}6.67& 2& 2\\ 2& 1.67& -3\\ 2& -3& -8.33\end{array}\right)$

and

${\mathit{\epsilon }}^{\prime }=\left(\begin{array}{ccc}0.1667-0.1499& 0.0467& 0.0467\\ 0.0467& 0.0500-0.1499& -0.0700\\ 0.0467& -0.0700& -0.1833-0.1499\end{array}\right)=\left(\begin{array}{ccc}0.0168& 0.0467& 0.0467\\ 0.0467& 0.0999& -0.0700\\ 0.0467& -0.0700& -0.3332\end{array}\right)$

Example 9

Problem Statement:

Within linear, homogeneous, isotropic elasticity theory, for a given strain state determine the stress state given the bulk modulus of 100 GPa and Lam$\acute{e}$ parameter of 50 GPa. Identify the hydrostatic stress, deviatoric stresses, strain dilatation, and strain deviator.

$\mathit{\epsilon }=\left(\begin{array}{ccc}0.002& 0.002& 0.001\\ 0.002& 0.002& 0.001\\ 0.001& 0.001& -0.001\end{array}\right)$

Solution:

We are given the strain, so returning a solution requires determining the stress, which can be determined by

${\sigma }_{ij}=\frac{E}{1+\nu }{\epsilon }_{ij}-\frac{\nu E}{(1+\nu )(1-2\nu )}{\epsilon }_{kk}{\delta }_{ij}$

where $\frac{\nu E}{(1+\nu )(1-2\nu )}$ is the Lamé parameter $\lambda$. Since

$G=\frac{E}{2(1+\nu )}$

and

$G=\frac{3(K-\lambda )}{2}$

we know

$\frac{E}{(1+\nu )}=3(K-\lambda )$

The stress expression becomes

${\sigma }_{ij}=3(K-\lambda ){\epsilon }_{ij}+\lambda {\epsilon }_{kk}{\delta }_{ij}$

Substituting and solving yields

$\mathit{\sigma }=\left(\begin{array}{ccc}0.45& 0.3& 0.15\\ 0.3& 0.45& 0.15\\ 0.15& 0.15& 0.0\end{array}\right)$GPa

The hydrostatic stress is

${\sigma }_m=\frac{{\sigma }_{kk}}{3}=\frac{{\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33}}{3}=\frac{0.45+0.45+0.0}{3}=0.30\text{ GPa}$

The strain dilatation is

$\mathrm{\Delta }=(1+{\epsilon }_{11})(1+{\epsilon }_{22})(1+{\epsilon }_{33})-1=(1+0.002)(1+0.002)(1-0.001)-1=0.003$

which results in a mean strain of

${\epsilon }_m=\frac{\mathrm{\Delta }}{3}=\frac{0.003}{3}=0.001$

An interesting observation is that the strain dilation and hydrostatic (mean) stress are related by the bulk modulus ${\sigma }_m=K\mathrm{\Delta }$, which in this case is $0.3=100\times 0.003$. When performing calculations it is important to use known checkpoints such as these to validate your work.

The deviatoric stress and strain are then determined by subtracting the mean stress and strain from their respective tensors diagonal.

${\mathit{\sigma }}^{\prime }=\left(\begin{array}{ccc}0.45-0.30& 0.30& 0.15\\ 0.30& 0.45-0.30& 0.15\\ 0.15& 0.15& 0.00-0.30\end{array}\right)=\left(\begin{array}{ccc}0.15& 0.30& 0.15\\ 0.30& 0.15& 0.15\\ 0.15& 0.15& -0.30\end{array}\right)$

and

${\mathit{\epsilon }}^{\prime }=\left(\begin{array}{ccc}0.002-0.001& 0.002& 0.001\\ 0.002& 0.002-0.001& 0.001\\ 0.001& 0.001& -0.001-0.001\end{array}\right)=\left(\begin{array}{ccc}0.001& 0.002& 0.001\\ 0.002& 0.001& 0.001\\ 0.001& 0.001& -0.002\end{array}\right)$

Example 10

Problem Statement:

For a single crystal of cubic zirconia, which has a elastic constants approximately $c_{11}=575$, $c_{12}=115$, and $c_{44}=75$ GPa, determine the elastic energy required to apply a uniaxial strain of 0.001 in the $\left[100\right]$ direction. Determine the elastic energy required to apply a uniaxial strain of 0.001 in the $\left[110\right]$ direction. Compute the Zener anisotropy ratio. Using isotropic elasticity theory, the elastic modulus of 200 GPa, and Poisson ratio of 0.3 the elastic modulus of $200GPa$ and the Poisson's ratio of $0.3$ to compute the elastic energy to apply a uniaxial strain of 0.001.

Solution:

Say that the $\left[100\right]$ is in the $x_1$ direction. The elastic energy, $U_o$, is

$U_o=\frac{1}{2}{\sigma }_{ij}{\epsilon }_{ij}$

so

$d{U}_o=\frac{1}{2}{\sigma }_{ij}d{\epsilon }_{ij}$

In the case of anisotropic elasticity theory

${\sigma }_{ij}=c_{ijkl}{\epsilon }_{kl}.$

In this case all ${\epsilon }_{ij}=0$ except ${\epsilon }_{11}=0.001$. Therefore all ${\sigma }_{ij}=0$ except ${\sigma }_{11}=c_{1111}{\epsilon }_{11}$, ${\sigma }_{22}=c_{1122}{\epsilon }_{22}$, and ${\sigma }_{33}=c_{1122}{\epsilon }_{33}$.

Substituting into the above equationsː

$\begin{array}{cc}d{U}_0& ={\epsilon }_{11}{c}_{1111}d{\epsilon }_{11}+\underset{=0uponintegration}{\underbrace{{\epsilon }_{11}{c}_{1122}d{\epsilon }_{22}+{\epsilon }_{11}{c}_{1122}d{\epsilon }_{33}}}\\ & ={\epsilon }_{11}{c}_{1111}d{\epsilon }_{11}\\ \\ U_0& =\underset{0}{\overset{{\epsilon }_{11}=0.001}{\int }}{\epsilon }_{11}{c}_{11}d{\epsilon }_{11}\\ & ={c_{11}\frac{{\epsilon }_{11}^2}{2}|}_0^{0.001}\\ & =(575*10^{9}Pa)\frac{0.001^2}{2}\\ & =288kJ\end{array}$

Example 11

Problem Statement:

For a single crystal of cubic zirconia, which has a elastic constants approximately $c_{11}=575$, $c_{12}=115$, and $c_{44}=75$ GPa, determine the elastic energy required to apply a uniaxial strain of 0.001 in the $\left[100\right]$ direction followed by a shear strain of 0.001 on the $\left(001\right)$ face in the $\left[010\right]$ direction.

Solution:

Obeying the given order of operation, first we apply a uniaxial load of $0.001$ in the $[100]$ direction. (Also covered in Example 10)

Note that our general equations are $dU={\sigma }_{ij}d{\epsilon }_{ij}$, and ${\sigma }_{ij}=c_{ijkl}{\epsilon }_{kl}$.

Restating the strain tensorː

$\epsilon =\left(\begin{array}{ccc}{\epsilon }_{11}& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$

From this we get the non-zero stress tensorː

$\sigma =\left(\begin{array}{ccc}{\epsilon }_{11}{c}_{11}& 0& 0\\ 0& {\epsilon }_{11}{c}_{12}& 0\\ 0& 0& {\epsilon }_{11}{c}_{12}\end{array}\right)$

Thus the only non-zero $d{\epsilon }_{ij}$ is $d{\epsilon }_{11}$, and by integrating our equation for $dU$ we getː $U_1=288kJ$

Now let's apply the shear $0.001$ on the $(100)$ face in the $[010]$ direction.

Our new stress tensor isː

$\sigma =\left(\begin{array}{ccc}{\epsilon }_{11}{c}_{11}& {\epsilon }_{12}{c}_{44}& 0\\ {\epsilon }_{12}{c}_{44}& {\epsilon }_{11}{c}_{12}& 0\\ 0& 0& {\epsilon }_{11}{c}_{12}\end{array}\right)$

Here, the only non-zero terms are $d{\epsilon }_{12}$, and $d{\epsilon }_{21}$. As these are equivalent terms, due to symmetry, we can solve for one and multiply the answer by two.

Once again utilizing our basic energy equation we getː

$\begin{array}{cc}dU& ={\sigma }_{ij}d{\epsilon }_{ij}={\epsilon }_{12}{c}_{44}d{\epsilon }_{12}\\ \\ U_2& =2\underset{0}{\overset{{\epsilon }_{12}}{\int }}{\epsilon }_{12}{c}_{44}d{\epsilon }_{12}\\ & =2{\left[c_{44}\frac{{\epsilon }_{12}^2}{2}\right]}_0^{{\epsilon }_{12}}\\ & =(75*10^{9}Pa)0.001^2\\ U_2& =75kJ\end{array}$

Finally, adding $U_1$ and $U_2$ together to get the total energy of this combined transformation gives us a final answer of $363kJ$.

Example 12

Problem Statement:

For a polycrystal specimen of cubic zirconia, which has a elastic constants approximately $c_{11}=575$, $c_{12}=115$, and $c_{44}=75$ GPa, use isotropic elasticity theory and the elastic modulus of $200GPa$ and the Poisson's ratio of $0.3$ to compute the elastic energy to apply a strain state

$\mathit{\epsilon }=\left(\begin{array}{ccc}0.001& 0.001& 0.000\\ 0.001& 0.000& 0.000\\ 0.000& 0.000& 0.000\end{array}\right)$

If the polycrystal material has a porosity of 2% approximately how much will this change the elastic modulus? Approximately how much will this change the elastic energy for this applied strain?

Solution:

Note that our general equations are $dU={\sigma }_{ij}d{\epsilon }_{ij}$, and ${\sigma }_{ij}=c_{ijkl}{\epsilon }_{kl}$.

Here, we can put the stress in terms of the elastic modulus and the Poisson's ratioː

${\sigma }_{ij}=\frac{E}{1+\nu }{\epsilon }_{ij}+\frac{\nu E}{(1+\nu )(1-2\nu )}{\epsilon }_{kk}{\partial }_{ij}$

Where the latter constant is equivalent to the Lamé Constant ($\lambda$)ː $\lambda =\frac{\nu E}{(1+\nu )(1-2\nu )}$

Briefly solving for the Lamé Constant yields usː

$\begin{array}{cc}\lambda & =\frac{\nu E}{(1+\nu )(1-2\nu )}\\ & =\frac{0.3*200GPa}{(1+0.3)(1-2(0.3))}=115.3846GPa\end{array}$

Keeping in mind that ${\epsilon }_{12}$ equals ${\epsilon }_{21}$ thanks to symmetry, the non-zero stresses for this problem areː

$\sigma =\left(\begin{array}{ccc}\frac{E}{1+\nu }{\epsilon }_{11}+\lambda {\epsilon }_{11}& \frac{E}{1+\nu }{\epsilon }_{12}& 0\\ \frac{E}{1+\nu }{\epsilon }_{12}& \lambda {\epsilon }_{11}& 0\\ 0& 0& \lambda {\epsilon }_{11}\end{array}\right)$

Therefore, we can write out the non-zero energy terms asː

$\begin{array}{cc}dU& ={\sigma }_{ij}d{\epsilon }_{ij}\\ & =(\frac{E}{1+\nu }+\lambda ){\epsilon }_{11}\mathrm{d}{\epsilon }_{11}+(\frac{E}{1+\nu }+\lambda ){\epsilon }_{12}\mathrm{d}{\epsilon }_{12}+(\frac{E}{1+\nu }+\lambda ){\epsilon }_{21}\mathrm{d}{\epsilon }_{21}+\lambda {\epsilon }_{11}\mathrm{d}{\epsilon }_{11}+\lambda {\epsilon }_{11}\mathrm{d}{\epsilon }_{11}\\ & =(\frac{E}{1+\nu }+\lambda ){\epsilon }_{11}\mathrm{d}{\epsilon }_{11}+2(\frac{E}{1+\nu }+\lambda ){\epsilon }_{12}\mathrm{d}{\epsilon }_{12}\\ \\ U& ={(\frac{E}{1+\nu }+\lambda )\frac{1}{2}{\epsilon }_{11}^2|}_0^{{\epsilon }_{11}}+{\left(\frac{E}{1+\nu }\right){\epsilon }_{12}^2|}_0^{{\epsilon }_{12}}\\ & =288kJ\end{array}$